I have some factoring questions that I wasn't sure how to do. The methods we can ue are common factoring, difference of squares, trinomial factoring, grouping, sum and difference of cubes, factor theorem, and extended factor theorem.) I'm not sure how to factor a trinomial though. These questions are to be factors completely as possible. If I could receive help it would be greatly appreciated.

1: x - d + (x-d)^2 I am thinking that this is simplified and can no longer be factored?

2: 36(2x-y)^2 - 25(u-2y)^2After expanding and simplifying ( I'm not sure if I should expand or not) I end up with 144x^2-144xy-25u^2+100uy-99y^2 and then I factored the 144 out and got:144(x^2-xy)-25u^2+100uy-99y^2 (not sure if factoring 144 out was correct and not sure what do to after this step)

3: Not sure how to factor this: y^5 - y^4 + y^3 - y^2 + y - 1

4: 9(x + 2y + z)^2 - 16(x - 2y + z) ^2 after expanding and simplifying I end up with -7x^2 + 100xy - 14xz - 28y^2 + 100xy - 7z^2 Then I factored some number out and get -7(x^2 + Z^2) + 100y(x + z) - 14(xz+y^2) ( I am not sure if this is correct and what to do next)

5: 3(x + 2W)^3 - 3p^3r^3 I get 3(x+2w)^3 - (pr) (^2r^2) not sure if this is right.

6: I am to common factor this question but am not sure how to:-3(x-1)^2 - 12(x-1)^-3

They may have mentioned a formula related to dividing by y - 1...? If they didn't give you a formula, then I don't see how you're supposed to "know" how to do this on your own, but:

. . . . .y4(y - 1) + y2(y - 1) + 1(y - 1)

Or, if you've learned how to do polynomial long division, you can divide the original polynomial by y - 1.

lia wrote:4: 9(x + 2y + z)^2 - 16(x - 2y + z) ^2

This is like (2) above: a difference of squares.

. . . . .[3(x +2y + z)]2 - [4(x - 2y + z)]2

. . . . .[3x + 6y + 3z]2 - [4x - 8y + 4z]2

You'll be able to simplify inside the factors.

lia wrote:5: 3(x + 2W)^3 - 3p^3r^3

You've got a common factor of 3, which leaves behind a difference of cubes:

. . . . .3[(x + 2W)3 - p3r3]

. . . . .3[(x + 2W)3 - (pr)3]

If they'd given you a3 - b3, you'd have factored as (a - b)(a2 + ab + b2). Apply the formula using x + 2W and pq instead of a and b. There won't be any simplifying that can be done inside the factors this time, though.

lia wrote:6: I am to common factor this question but am not sure how to:-3(x-1)^2 - 12(x-1)^-3

I'm not sure what is meant by "common factoring" something, and the negative exponent on the x - 1 in the second term would seem to make this impossible...?

2) You have the formula a2 - b2 = (a - b)(a + b). Instead of "a" and "b" for the difference of squares, you have "12x - 6y" and "5u - 10y". But the pattern is the same: ([first thing] - [second thing])([first thing] + [second thing]).

4) To "simplify", you just combine the "like" terms (the terms containing "y").

5) I'm not sure what you're doing on this exercise...? You don't have (x)3 - (2W)3; you have (x + 2W)3 - (pr)3. Try copying down the differences-of-cubes formula, a3 - b3 = (a - b)(a2 + ab + b2), and then replacing "a" with "x - 2W" and "b" with "pq". See where that leads.